Killing form is negative definite

Question

Let $G$ be a compact connected semisimple Lie group and $\frak g$ its Lie algebra. It is known that the Killing form of $\frak g$ is negative definite. What about the Killing form $B$ of the complex semisimple Lie algebra ${\frak g}_{\Bbb C}={\frak g}\otimes\Bbb C$?

In particular:

If $X\in{\frak g}_{\Bbb C}$ lies in the Cartan subalgebra ${\frak h}_{\Bbb C}\subseteq{\frak g}_{\Bbb C}$ and $B(X,X)=0$, does that imply that $X=0$?

Answer 1

Take the compact simple real Lie algebra $\mathfrak{su}(2)$. The Killing form is negative definite. Its complexification is isomorphic to the Lie algebra $\mathfrak{sl}_2(\mathbb{C})$, where the Killing form is non-degenerate, but does not satisfy $B(x,x)=0$ implies $x=0$. If $x,y,h$ is the standard basis, then $B(x,x)=0$.

Answer 2

The Killing form of $\mathfrak g_{\mathbb C}$ is complex bilinear by construction, so in any complex subspace of $\mathfrak g_{\mathbb C}$ you find non-zero vectors which are isotropic.